For instance, If there are three enemies involved in a battle over one region then there will be two attackers (pawns and knights will use attack values) and one defender.

In this case the defender has to contend with both enemies combined attack score in the first round(s) of fighting. If the attackers defeat the defender they then turn on one another.

The RED ARMY and the BLUE ARMY attack the WHITE ARMY in the same round. All three players are enemies.

Attacking RED has 15k
Attacking BLUE 12p
Defending WHITE 13p

(yes in this example BLUE is very silly)

RED's attack of 15k at 5/6 = 75/6 = 12 kills and another at 3/6 (%50)

BLUE's attack 12p at 1/6 = 12/6 = 2 kills against the defender.

We can safely say WHITE is dead at this point, but it will take a few attackers with it on the way.

WHITE's defence of 13p at 3/6 = 39/6 = 6 kills and another at 3/6 (%50)

This is where I come unstuck. I'm not sure how Vox decides who of the two attackers sustains the damage, but it appears to give all the damage to BLUE so that in the 2nd round RED has 15k and BLUE has 6p or 5p (if WHITE rolls a 4-6 BLUE loses another troop)

If this result is true I'd love to know the reasons for the imbalance in damage sustained. I'm facing just such an attacking move in a game now...

All 12 defeding paws will die in the first round due to RED's knight force, but the 12 defenders will kill 6 attackers. What attacking units will die in this case?
It is obvious that pawns need to die before knights, so the result will be:
RED wins 0p/14k left. (I hope so, never tested it )

All 14 defending pawns will again die in the first round (at least 12p due to RED's knights and 2p due to 12p attacking pawns), but what attacking 7 pawns will die now?
Honestly, I am not sure what will happen. Possible solutions are
1) A pawn stack will be randomly selected and 6p will be reduced from that one. Then the last 1p will be reduced from the other one.
2) Losses evenly split to both pawn stacks (4:3 in that case)
If 1) is true and selects BLUE's stack first, the battle will be over. In any other case there will be another battle round between the two attackers.

drSnuggles wrote:
Honestly, I am not sure what will happen. Possible solutions are
1) A pawn stack will be randomly selected and 6p will be reduced from that one. Then the last 1p will be reduced from the other one.
2) Losses evenly split to both pawn stacks (4:3 in that case)
If 1) is true and selects BLUE's stack first, the battle will be over. In any other case there will be another battle round between the two attackers.

Well, hmm, dustin would have to comment to clarify.

It appears 2) is the answer. 4:3 in the example you've used. Good news imo, it's the most logical and fairest outcome.